Question 14 - Miscellaneous - Chapter 8 Class 11 Sequences and Series
Last updated at April 16, 2024 by Teachoo
Miscellaneous
Misc 2
Misc 3
Misc 4 Important
Misc 5
Misc 6
Misc 7 Important
Misc 8
Misc 9
Misc 10 Important
Misc 11 (i) Important
Misc 11 (ii)
Misc 12 Important
Misc 13
Misc 14 Important
Misc 15 Important
Misc 16
Misc 17 Important
Misc 18 Important
Question 1
Question 2
Question 3 Important
Question 4
Question 5
Question 6 Important
Question 7
Question 8
Question 9 Important
Question 10
Question 11 Important
Question 12
Question 13 Important
Question 14 You are here
Miscellaneous
Last updated at April 16, 2024 by Teachoo
Misc 26 Show that (1 22 + 2 32 + + n (n + 1)2)/(12 2 + 22 3 + + n2 (n + 1)) = (3n + 5)/(3n + 1) Taking L.H.S (1 22 + 2 32 + + n (n + 1)2)/(12 2 + 22 3 + + n2 (n + 1)) We solve denominator & numerator separately Solving numerator Let numerator be S1 = 1 22 + 2 32 + + n (n + 1)2 nth term is n (n + 1)2 Let an = n(n + 1)2 = n(n2 + 1 + 2n) = n3 + n + 2n2 Now finding S1 = (( ( + 1))/2)^2 + 2(( ( +1)(2 +1))/6) + n(n+1)/2 = ( ( + 1))/2 (n(n+1)/2 " + " (2(2 +1))/3 " + 1" ) = ( ( + 1))/2 (( 3 ( +1) + 2 2(2 +1)+ 6)/6) = (n(n + 1))/(2 6)[3n(n + 1) + 4(2n + 1) + 6] = (n(n + 1))/12[3n2 + 3n + 8n + 4 + 6] = ( ( + 1))/12[3n2 + 11n + 10] = ( ( + 1))/12[3n2 + 5n + 6n + 10] = ( ( + 1))/12[n(3n + 5) + 2(3n + 5)] = ( ( + 1))/12[(n + 2)(3n + 5)] Thus, S1 = ( ( + 1))/12[(n + 2)(3n + 5)] Now solving denominator Let denominator be S2 = 12 2 + 22 3 + + n2 (n + 1) nth term is n2(n + 1) Let bn = n2(n + 1) bn = n3 + n2 Now, calculating S2 = (( ( + 1))/2)^2 + (( ( +1)(2 +1))/6) = ( ( + 1))/2 (n(n+1)/2 " + " ((2 +1))/3) = ( ( + 1))/2 (n(n+1)/2 " + " ((2 +1))/3) = ( ( + 1))/2 (( 3 ( +1) + 2 (2 +1))/6) = (n(n + 1))/(2 6) (3n(n + 1) + 2(2n + 1)) = (n(n + 1))/12 (3n2 + 3n + 2(2n + 1)) = (n(n + 1))/12 (3n2 + 3n + 4n + 2) = (n(n+1))/12 (3n2 + 7n +2) = (n(n+1))/12 (3n2 + 6n + n +2) = (n(n+1))/12 (3n(n + 2) + 1(n +2)) = (n(n+1)(n+2)(3n+1))/12 Thus, S2 = (n(n+1)(n+2)(3n+1))/12 Now, Taking L.H.S (1 22 + 2 32 + + n (n + 1)2)/(12 2 + 22 3 + + n2 (n + 1)) = 1/ 2 = ((n(n+1)(n+2)(3n+5))/12)/((n(n+1)(n+2)(3n+1))/12) = (n(n+1)(n+2)(3n+5))/12 12/(n(n+1)(n+2)(3n+1)) = (n(n+1)(n+2)(3n+5))/(n(n+1)(n+2)(3n+1)) = ((3n+5))/((3n+1)) = R.H.S Hence L.H.S = R.H.S Hence proved.